Question

Determine the scalar product of the functions \(f(x)=\sin (2 n \pi x / L)\) and \(g(x)=\cos (2 p \pi x / L)\) on the domain \(0 \leq x \leq L\), where \(n \neq p\) are positive integers greater than zero.

Short answer

Answer: The scalar product of the given functions is 0.

Step-by-step solution

Identify the functions f(x) and g(x)

The given functions are \(f(x)=\sin (2 n \pi x / L)\) and \(g(x)=\cos (2 p \pi x / L)\), where n and p are positive integers greater than zero and not equal to each other.

Write the integral for the scalar product

To find the scalar product of the given functions, we will integrate the product of f(x) and g(x) over the domain \(0 \leq x \leq L\). This can be represented as: $$\langle f, g \rangle = \int_{0}^{L} f(x)g(x) dx = \int_{0}^{L} \sin (2 n \pi x / L) \cos (2 p \pi x / L) dx$$

Use the product-to-sum identity

To facilitate the integration, we can use the product-to-sum identity for sine and cosine functions: $$\sin A \cos B = \frac{1}{2}(\sin(A - B) + \sin(A + B))$$ In our case, \(A = 2 n \pi x / L\) and \(B = 2 p \pi x / L\). Applying the identity, we get: $$\sin (2 n \pi x / L) \cos (2 p \pi x / L) = \frac{1}{2}(\sin(2 (n - p) \pi x / L) + \sin(2 (n + p) \pi x / L))$$

Substitute the product-to-sum identity into the integral

We will now substitute the product-to-sum identity result into our integral and simplify: $$\langle f, g \rangle = \int_{0}^{L} \frac{1}{2} (\sin(2 (n - p) \pi x / L) + \sin(2 (n + p) \pi x / L)) dx$$

Solve the integral

Now, we integrate the sum term by term: $$\langle f, g \rangle = \frac{1}{2} \int_{0}^{L} \sin(2 (n - p) \pi x / L) dx + \frac{1}{2} \int_{0}^{L} \sin(2 (n + p) \pi x / L) dx$$ Recall the integral of sine functions: $$\int \sin(ax) dx = -\frac{1}{a} \cos(ax) + C$$ Using this result, we get: $$\langle f, g \rangle = -\frac{L}{4 \pi(n - p)}\left[\cos(2 (n - p) \pi x / L)\right]_{0}^{L}+\frac{L}{4 \pi(n + p)}\left[\cos(2 (n + p) \pi x / L)\right]_{0}^{L}$$

Evaluate the integral

Now, we evaluate the integral. We know that \(\cos(2 \pi kx) = 1\) if k is an integer, and plugging in the limits \(x = L\) and \(x = 0\): $$\langle f, g \rangle = -\frac{L}{4 \pi(n - p)}(\cos(2 (n - p) \pi) - \cos(0)) + \frac{L}{4 \pi(n + p)}(\cos(2 (n + p) \pi) - \cos(0))$$ Since \(\cos(2 (n \pm p) \pi) = 1\) and \(\cos(0) = 1\), we have: $$\langle f, g \rangle = -\frac{L}{4 \pi(n - p)}(1 - 1) + \frac{L}{4 \pi(n + p)}(1 - 1) = 0$$ Thus, the scalar product of the given functions f(x) and g(x) is 0.